[proofplan] We code the mixing Axiom A basic set by a mixing subshift of finite type using a Markov partition. On the symbolic system, the Parry measure is the unique measure of maximal entropy. The finite-to-one coding preserves entropy for invariant lifts and preserves topological entropy, so every maximal entropy measure on $\Lambda$ lifts to a symbolic maximal entropy measure. Uniqueness on the symbolic side then forces uniqueness on $\Lambda$. [/proofplan] [step:Choose a finite-to-one mixing Markov coding of the basic set] Let \begin{align*} T: \Lambda \to \Lambda,\quad x \mapsto f(x) \end{align*} denote the restricted map. By the Markov partition theorem for mixing Axiom A basic sets (citing a result not yet in the wiki: Markov partition for Axiom A basic sets), there exist a finite alphabet subshift of finite type \begin{align*} \sigma: \Sigma_A \to \Sigma_A \end{align*} with topologically mixing transition matrix $A$, and a continuous surjective finite-to-one map \begin{align*} \pi: \Sigma_A \to \Lambda \end{align*} such that \begin{align*} \pi \circ \sigma = T \circ \pi. \end{align*} Thus $\pi$ is a factor map from the symbolic system $(\Sigma_A,\sigma)$ onto $(\Lambda,T)$. [/step] [step:Push the Parry measure forward to obtain an invariant measure on $\Lambda$] Let $m_A$ be the Parry measure on $\Sigma_A$. Define the Borel probability measure \begin{align*} \mu_B := \pi_*m_A \end{align*} on $\Lambda$ by $\mu_B(E) = m_A(\pi^{-1}(E))$ for every Borel set $E \subset \Lambda$. Since $m_A$ is $\sigma$-invariant and $\pi \circ \sigma = T \circ \pi$, the pushforward measure $\mu_B$ is $T$-invariant. Indeed, for every Borel set $E \subset \Lambda$, \begin{align*} \mu_B(T^{-1}(E)) = m_A(\pi^{-1}(T^{-1}(E))). \end{align*} Using the semiconjugacy relation gives \begin{align*} \pi^{-1}(T^{-1}(E)) = \sigma^{-1}(\pi^{-1}(E)). \end{align*} Therefore, by $\sigma$-invariance of $m_A$, \begin{align*} \mu_B(T^{-1}(E)) = m_A(\sigma^{-1}(\pi^{-1}(E))) = m_A(\pi^{-1}(E)) = \mu_B(E). \end{align*} [/step] [step:Show that the pushed-forward Parry measure has topological entropy] The [Parry measure theorem](/theorems/6793) for mixing subshifts of finite type (citing a result not yet in the wiki: Parry measure is the unique measure of maximal entropy for a mixing subshift of finite type) gives \begin{align*} h_{m_A}(\sigma) = h_{\mathrm{top}}(\sigma). \end{align*} Since $\pi: \Sigma_A \to \Lambda$ is a finite-to-one factor map coming from a Markov partition, the symbolic coding preserves topological entropy: \begin{align*} h_{\mathrm{top}}(\sigma) = h_{\mathrm{top}}(T). \end{align*} Also, finite-to-one factor maps have zero relative entropy over the factor for invariant measures (citing a result not yet in the wiki: finite-to-one factor maps preserve measure entropy). Applying this to $m_A$ and $\mu_B=\pi_*m_A$ gives \begin{align*} h_{m_A}(\sigma) = h_{\mu_B}(T). \end{align*} Combining the three equalities yields \begin{align*} h_{\mu_B}(T) = h_{\mathrm{top}}(T). \end{align*} Thus $\mu_B$ is a measure of maximal entropy for $T$. [guided] The purpose of this step is to transfer maximality from the symbolic model to the basic set. The symbolic side is where the measure is explicitly known: the Parry measure theorem for mixing subshifts of finite type states that the Parry measure $m_A$ is $\sigma$-invariant and satisfies \begin{align*} h_{m_A}(\sigma) = h_{\mathrm{top}}(\sigma). \end{align*} The hypotheses of that theorem are met because $\Sigma_A$ was chosen from a Markov partition whose transition matrix $A$ is topologically mixing. Next, the Markov coding is finite-to-one and onto, so it is not merely an arbitrary factor map. The standard entropy theorem for finite-to-one Markov codings says that the symbolic model and the basic set have the same topological entropy: \begin{align*} h_{\mathrm{top}}(\sigma) = h_{\mathrm{top}}(T). \end{align*} This equality is essential: if the coding had extra topological entropy, the symbolic maximal measure could project to a measure whose entropy is too small on $\Lambda$. Finally, we compare measure entropies. The factor relation $\pi \circ \sigma = T \circ \pi$ always gives a factor system, and for a finite-to-one factor map the conditional entropy along the fibres is zero for invariant measures. Applied to $m_A$ and its pushforward $\mu_B=\pi_*m_A$, this gives \begin{align*} h_{m_A}(\sigma) = h_{\mu_B}(T). \end{align*} Combining the three displayed equalities gives \begin{align*} h_{\mu_B}(T) = h_{m_A}(\sigma) = h_{\mathrm{top}}(\sigma) = h_{\mathrm{top}}(T). \end{align*} Therefore the pushed-forward Parry measure $\mu_B$ attains the topological entropy of $T$, so it is a measure of maximal entropy on $\Lambda$. [/guided] [/step] [step:Lift an arbitrary invariant measure on $\Lambda$ to the symbolic system] Let $\nu$ be any $T$-invariant Borel probability measure on $\Lambda$. By invariant measure lifting for factor maps (citing a result not yet in the wiki: invariant measure lifting for factor maps), applied to the continuous surjective factor map $\pi: \Sigma_A \to \Lambda$, there exists a $\sigma$-invariant Borel probability measure $\tilde{\nu}$ on $\Sigma_A$ such that \begin{align*} \pi_*\tilde{\nu} = \nu. \end{align*} Since $\pi$ is finite-to-one, the relative entropy of $\tilde{\nu}$ over $\nu$ is zero. Hence the finite-to-one entropy preservation theorem gives \begin{align*} h_{\tilde{\nu}}(\sigma) = h_\nu(T). \end{align*} [/step] [step:Use symbolic uniqueness to force uniqueness on the basic set] Assume that $\nu$ is a $T$-invariant Borel probability measure on $\Lambda$ with \begin{align*} h_\nu(T) = h_{\mathrm{top}}(T). \end{align*} Choose a $\sigma$-invariant lift $\tilde{\nu}$ of $\nu$ as in the previous step. Then \begin{align*} h_{\tilde{\nu}}(\sigma) = h_\nu(T) = h_{\mathrm{top}}(T) = h_{\mathrm{top}}(\sigma). \end{align*} Thus $\tilde{\nu}$ is a measure of maximal entropy for the mixing subshift of finite type $(\Sigma_A,\sigma)$. By uniqueness of the Parry measure for a mixing subshift of finite type, $\tilde{\nu}=m_A$. Pushing forward by $\pi$ gives \begin{align*} \nu = \pi_*\tilde{\nu} = \pi_*m_A = \mu_B. \end{align*} Therefore every $T$-invariant Borel probability measure with entropy $h_{\mathrm{top}}(T)$ equals $\mu_B$. Since the previous step already proved that $\mu_B$ attains this entropy, $\mu_B$ is the unique measure of maximal entropy for $f|_\Lambda$. [/step]